JS如何对包含日期属性的对象数组按年份聚合，求和对应数值属性

核心实现逻辑 #

整体分为两步完成：

• 遍历原始数组，提取日期中的年份作为分组标识，累加同一年份的B字段值
• 将分组后的年份和对应总和，转换为指定结构的新数组

JavaScript 实现 #

const AA = [
  {A: '2016-01-01', B: 12},
  {A: '2016-02-01', B: 122},
  {A: '2017-03-01', B: 162},
  // 剩余原始数据
]

// 按年份累加B值
const yearMap = AA.reduce((result, current) => {
  const year = current.A.slice(0, 4)
  result[year] = (result[year] || 0) + current.B
  return result
}, {})

// 生成目标格式数组
const newAA = Object.keys(yearMap).map(year => ({
  A: `${year}-01-01`,
  B: yearMap[year]
}))

Python 实现 #

如果确定日期格式固定为YYYY-MM-DD，可以直接用字符串切片提取年份：

AA = [
    {"A": "2016-01-01", "B": 12},
    {"A": "2016-02-01", "B": 122},
    {"A": "2017-03-01", "B": 162},
    # 剩余原始数据
]

year_sum = {}
for item in AA:
    year = item["A"][:4]
    year_sum[year] = year_sum.get(year, 0) + item["B"]

newAA = [{"A": f"{year}-01-01", "B": sum_val} for year, sum_val in year_sum.items()]

如果日期格式存在波动，建议用日期解析工具提取年份，更稳妥：

from datetime import datetime

AA = [
    {"A": "2016-01-01", "B": 12},
    {"A": "2016-02-01", "B": 122},
    # 剩余原始数据
]

year_sum = {}
for item in AA:
    # 按实际日期格式调整strptime的第二个参数
    date = datetime.strptime(item["A"], "%Y-%m-%d")
    year = str(date.year)
    year_sum[year] = year_sum.get(year, 0) + item["B"]

newAA = [{"A": f"{year}-01-01", "B": sum_val} for year, sum_val in year_sum.items()]

内容的提问来源于stack exchange，提问作者geo